How liquid pressure = dgh can be generalized

I've seen the standard derivation of the expression for liquid pressure
P = dgh where,
d = density of the liquid;
g = acceleration due to gravity;
h = height of liquid column
in many text books has been done by using a specific example of a cylindrical vessel.
In such a case, the geometry of the cylinder allows us to write Volume = Base Area x Height
So, the above result is trivial.
But in many cases where such a direct link between volume of the vessel and its area and the other dimension doesn't exist, like for a cone, its volume is V = [itex]\pi[/itex] r[itex]^{2}[/itex] x h x (1/3) , we get that extra constant 1/3 and liquid pressure is no more dgh I think.
How then can a specific example be used to speak for the general?
I've seen this kind of easy generalization in many other parts of physics such as torque on a plane coil in a magnetic field([itex]\tau[/itex]=BINAcosx), where the actual derivation has been performed using a simple rectangular coil and said that the result is true for any shape of plane coil with the same area.
I understand that this has been done for simplifying things but is there some law in physics that I do not know of that allows for such bold assumptions to be made even just by watching patterns?

Staff: Mentor

Liquid pressure is a function of depth only, the geometry does not matter (in equilibrium).
While you have a different amount of water in a cone, you also have (vertical) pressure on the walls there. A cylinder is easier to calculate, as all walls are vertical.

I understand that this has been done for simplifying things but is there some law in physics that I do not know of that allows for such bold assumptions to be made even just by watching patterns?

Sometimes there is. In the example of the coil, you can use vector calculus to see the identity.

But why is there the constant (1/3) in case of the cone as opposed to that in the case of a cylinder? Even if a cone and a cylinder are of the same height and base area, the pressure at the base of cone seems to be lesser by one-third than in cylinder according to this way of looking at liquid pressures. What is wrong here?

Staff: Mentor

The pressure is the same in the cone.
As I said, if you want to do the same derivation there, you would have to consider pressure from the walls, which is quite messy. It is easier to verify that the fluid has to have the same pressure at the same depth, and look at the interior of the fluid. This way, you get p=dgh independent of the geometry.

A more sophisticated derivation which works for any shape of container uses the ideas of work and energy.

Imagine a small cylinder fitted with a piston of area A, which is inserted in the liquid so the piston is at depth h, at any orientation we choose, and in contact with the liquid on one side (of the piston).

If we push the piston out, a small distance x into the liquid, we'll displace a mass ρAx of liquid, and an equal mass of liquid will, effectively, be displaced from depth h to the surface, gaining gravitational potential energy ρAxhg.

We can equate this gain in GPE to the work p dV = p A dx done by the piston against the liquid pressure p.

This gives p = ρgh.

This is delightful because it shows the result to be totally independent of the container shape and of the orientation of the surface against which the liquid pushes. The piston is assumed to move slowly enough for viscous (dissipative) forces to be negligible; after all we are dealing with hydrostatics!

I first saw this derivation in A Second Course of Mechanics by A E E McKenzie, a highly respected writer of British school Physics textbooks of some fifty years ago.

in many text books has been done by using a specific example of a cylindrical vessel. In such a case, the geometry of the cylinder allows us to write Volume = Base Area x Height So, the above result is trivial.

Trivial? Hmm. You're assuming that the pressure on the base is uniform and that the net pressure from the sides of the container is zero. In the case of the cylinder, the assumptions hold. But in the case of a conical container, do they hold? No, that's why you can't use the same argument in that case.

There are several ways to generalise to find the pressure inside some arbitrary container. I think some people have already replied with good answers. Here is how I would do it. First, make some assumptions 1) the fluid is static (i.e. not moving) 2) there is zero shear forces within the fluid 3) the pressure in the fluid is the same in all direction (And therefore can be written as a scalar pressure) 4)gravity and density are constant with respect to time and space.

We can write the force on any arbitrary parcel of fluid as due to the force of pressure on the surface of that parcel, plus the force due to gravity on that parcel:
[tex]\vec{F} = - \oint P d \vec{A} - \int \rho g \hat{z} dV [/tex]
Where P is the scalar pressure, [itex]\rho[/itex] is the density of the fluid and g is the absolute value of gravity, i.e. 9.81 m/s^2 We also know that because of Gauss' law, we can write:
[tex]\oint P d \vec{A} = \int \nabla P dV [/tex]
So now we have:
[tex]\vec{F} = \int - (\nabla P) - \rho g \hat{z} dV[/tex]
We have assumed that the fluid is static. This implies that the force on any arbitrary parcel of fluid is equal to zero, so the integral on the right hand side must be zero, over any arbitrary volume. To achieve this, we must make the integrand equal to zero at all points in the fluid. Therefore:
[tex] \nabla P = - \rho g \hat{z} [/tex]
Now, since P is a scalar, we also have the equation:
[tex]dP = \nabla P \cdot d \vec{L} [/tex]
Where [itex]d \vec{L} = dx \hat{x} + dy \hat{y} + dz \hat{z} [/itex] So we can use this, along with our equation to get:
[tex]dP = - \rho g \hat{z} \cdot d \vec{L} [/tex]
And therefore, substituting in [itex]d \vec{L}[/itex] we get:
[tex]dP = - \rho g dz [/tex]
And since we have assumed that density and gravity are constant,
[tex]P = - \rho g z + C[/tex]
Where C is just a constant. If we now specify that h= -z (i.e. h measures distance downwards) and if we specify that C=0 when h=0, we finally get:
[tex]P= \rho g h [/tex]

That's why the bracket should, I think, be moved to between the [itex]\widehat{z}[/itex] and the dV, but this is the merest nitpick, and I wish I hadn't mentioned it. I thought your derivation was a tour de force, though the work-energy method delivers the result convincingly in a couple of lines with elementary mathematics - for someone willing to use a slightly less direct method.

yeah. I try to use work-energy method most of the time. But when I am not sure about something, I go back to this kind of method. I think that means I need to get more familiar with work-energy methods. Lagrangian mechanics and Hamiltonian vector fields, I have had an introduction to, but I'm still not as comfortable with.

Like for example, when the density and gravity are not constant, I know what to do with my method, since
[tex]\frac{dP}{dz} = - \rho g [/tex]
Then when [itex]\rho[/itex] and g are dependent on z, then you can just integrate with respect to z (where the limits are just the initial and final positions), then you will get the change in pressure between those two positions. But using the work-energy method, I would not be sure what to do.

Hadn't thought of extending the energy method to more difficult situations. Presumably its viability depends on whether or not ρg, in its dependency on position in the liquid, is derivable from a scalar potential function.

That has helped me realise something that I missed. I should have said another assumption is that the pressure at some fixed point does not vary with time.

When I used Gauss' law, [itex]\nabla P [/itex] must be the partial derivatives of the spatial dimensions, while holding time constant (If I'm correct... I think this is explained by the maths of differential forms). So therefore,
[tex]dP = \nabla P \cdot d \vec{L} + \frac{\partial P}{\partial t} dt [/tex]
We must have this extra term, because time is being held constant in the first term. So from here, if I assume that the second term is zero, we get [itex]dP = \nabla P \cdot d \vec{L}[/itex] and the rest of the derivation continues the same as I wrote above.

Instead, we could not introduce this assumption, and keep the extra term. So for time-dependent problems, it gets a bit more complicated. You'd probably have to use Reynolds transport theorem.